Let me try to make several statements which I believe to be true - I'm basically hoping that somebody will point out where I make an error (or errors).

1) per wikipedia, any two-level quantum mechanical system can be used as a qubit.

2) per wikipedia, an example of such a two-level QM system is given by $$ \psi(t) = M \psi(0), $$ where for all $t$ we have $\psi(t)\in \mathbb C^2$ and $M$ is the diagonal matrix $$ M := \begin{pmatrix} e^{i\omega t} & 0 \\ 0 & e^{-i\omega t} \end{pmatrix}, $$ with $\omega$ being a fixed non-zero real number.

3) when one talks about quantum computing, it is customary to use e.g. the notation $$|0\rangle =\begin{pmatrix}1 \\ 0 \end{pmatrix}$$ $$|1\rangle = \begin{pmatrix}0 \\ 1 \end{pmatrix}$$

4) It doesn't make sense to talk about this system being permanently in the state, say, $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$, because even if for some $t_0$ we have $$ \psi(t_0) = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) $$ then for all $t>t_0$ which are not of the form $\frac{n\pi}{\omega}$ we have that the vectors $\psi(t_0)$ and $\psi(t)$ are linearly independent, so they correspond to different states.

Update/Clarification: This seems to me to be a problem because of the following. Suppose that we want to measure in the basis consisting of the vectors $|+\rangle:= \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $|-\rangle := \frac{1}{\sqrt{2}}(|0\rangle -|1\rangle)$ . Suppose that $\omega = 2\pi$ and at $t=0$ we have $\psi(0)=|+⟩$. Then at $t=\frac{1}{4}$ we have $\psi(\frac{1}{4}) = \frac{i}{\sqrt{2}}(|0⟩−|1⟩)=i|−⟩$. It follows that at times $t=0$ and $t=\frac{1}{4}$ this "same" qubit would give different measurement outcomes with probability 1 - surely this is not how a quantum computer is supposed to work?

5) So it would seem that such a system can be used for quantum computing only under the assumption that all operations in the quantum computer are done only precisely at the times of the form $t = n\pi/\omega$ - is this right? (I know very little about engineering - but this seems to be a difficult engineering obstacle - is it not?)

6) A different idea is that, contrary to what wikipedia says, perhaps not any two-level QM system can be used as a qubit, but rather only those where the energy is the same for all states? Then it would follow that the evolution is given by a diagonal matrix with constant coefficients.

(I have some follow-up questions, but they depend on what answers I'll get).


3 Answers 3


Let me argue by analogy:

We know that we can draw a straight line on a sheet of paper by taking a pen and moving it from left to right.

Now imagine the paper is rotating. If we try to follow the same procedure as before to draw a line, we will instead be left with a spiral-shaped curve.

Following your reasoning, this would mean that we cannot draw a straight line on a rotating paper - or, at the very least, we can only move our hand at points in time where the paper has rotated by a multiple of $2\pi$.

This is of course not true -- what we have to do is to adapt the way in which we move our hand, taking into account the rotation of the paper.

The same is true for performing quantum gates: Since we know the energy difference of the qubits and thus the rotation $e^{i\omega t}$, what we have to do is to take this rotation into account and carry out our quantum gates in suitably adapted way.

Effectively, this is very close to working in the Heisenberg picture (or interaction picture) rather than the Schrödinger picture, where we consider our operations and measurements to rotate, rather than the state of the system.

  • $\begingroup$ if anything, it seems a mix between the Heisenberg picture and the Schroedinger picture, since the actual unitary gates do change the state vector. In any case, could you give a concrete example where the gates evolve in time? In photon-based pop-sci representations of quantum computers gates consist of "static" objects, such as beam splitters, etc. (as I tried to argue in the question, I'm guessing that this is sort of expected, as in such a photon-based quantum computer all states have the same energy) $\endgroup$ Oct 5, 2019 at 4:26
  • $\begingroup$ @ŁukaszGrabowski Indeed, it is the interaction picture. -- I don't know details of implementations off the top of my head. It would make sense to ask a new question. Your question was whether this makes QC impossible, and the answer is clearly "no". $\endgroup$ Oct 5, 2019 at 8:15
  • $\begingroup$ Of course, experimentally our knowledge of $\omega$ will never be exact, but in any viable QC implementation it will be measurable to very high accuracy. $\endgroup$
    – Rococo
    Jan 11, 2020 at 20:59

These questions are a bit hard to answer in full generality, as quantum architectures can differ quite a bit in the way the quantum information is encoded and processed. But generally speaking, yes, it is true that most quantum systems/qubits will have some dephasing time due to environmental noise etc. Similarly, very often operations will have to be timed accurately in order to use the qubits before the information is degraded. How hard this actually is fully depends on the architecture though. For example, a qubit encoded in the polarisation of a single photon in clear air will have relatively long decoherence times.

Moreover, when dephasing is a problem, there are ways to overcome it and "renew" the information before it is too degraded. You might want to have a look at things like quantum error correction and quantum repeaters about this.

  • $\begingroup$ I found this video helpful: youtube.com/watch?v=w7398u8G588 I find it hard to "look at things like quantum error correction", when all the actual quantum devices are so extremely far away from the idealised theoretical description of quantum computing - it's even hard to say where do you actually start to error-correct as pretty much every single aspect introduces errors. The kind of errors I'm asking about I beleive are called "coherent errors" in the above talk, where the problem is e.g. that you rotate the system slightly too much or too little. This is my "impression" anyway... $\endgroup$ Sep 30, 2019 at 12:11
  • 1
    $\begingroup$ @glS Your title edit seems to have little to do with the question, which seems about rotating phases rather than dephasing. $\endgroup$ Oct 3, 2019 at 23:27
  • $\begingroup$ @NorbertSchuch you are right, your understanding of the question is more on point $\endgroup$
    – glS
    Oct 3, 2019 at 23:37
  • $\begingroup$ I think this answer also misunderstood the question being asked, so while it is not strictly incorrect I give it -1 for being more likely to only confuse the OP further. $\endgroup$
    – Rococo
    Jan 11, 2020 at 21:01

A couple of comments which may make things clearer:

(a) We can never measure phase in QM, so we don't care what value $\omega t$ takes - it does not affect the outcome of any experiment or measurement or observation. It is only amplitudes that we care about. So a system that is in state

$\psi(t) = \frac{1}{\sqrt{2}}(e^{i \omega t}|0\rangle + e^{-i \omega t}|1\rangle)$

has the same stochastic behavior when you measure observable $\{|0\rangle, |1\rangle\}$ not matter what time $t$ we observe it - only the amplitudes of its components matter.

(b) Although the simplest quantum computing system could contain only one qubit, it wouldn't do anything useful or interesting. Any useful or interesting system will contain multiple qubits, which we want to be in an entangled state i.e. the overall state of the system cannot be expressed as the product of the states of individual qubits. The biggest engineering challenge of quantum computing is keeping the qubits in an entangled state for long enough to do something useful with them. So thinking of a quantum computer in terms of the behaviour of individual qubits is probably not a good approach - it is the entangled state of the whole collection of qubits that is important.

  • $\begingroup$ I'm not interested in b) for now. As for a), what happens if you measure in the basis consisting of the vectors $|+\rangle := \frac{1}{\sqrt{2}}( |0\rangle + |1\rangle)$ and $|-\rangle := \frac{1}{\sqrt{2}}( |0\rangle - |1\rangle)$ ? Suppose that $\omega = 2\pi$ and at $t=0$ we have $\psi(0) = |+\rangle$. Then at $t=\frac{1}{4}$ we have $\psi(\frac{1}{4}) = \frac{i}{\sqrt{2}}(|0\rangle -|1\rangle) = i|-\rangle$. It follows that at times $t=0$ and $t=\frac{1}{4}$ this "same" state would give different measurement outcomes with probability $1$. $\endgroup$ Sep 25, 2019 at 11:06
  • $\begingroup$ I've added the comment above as a clarifying paragraph to point 4) in the question. $\endgroup$ Sep 25, 2019 at 11:41
  • $\begingroup$ I don't agree with statement $a$ (unless I misunderstood it). The state at $t=0$ is $|+\rangle$ which is distinguishable from the state at $t= \frac{\pi}{2\omega}$ which is $i|-\rangle$. States that differ only by a global phase are indistinguishable. $\endgroup$ Sep 25, 2019 at 11:42
  • $\begingroup$ @ŁukaszGrabowski You have changed what you are measuring. You can have a state where probabilities are time independent when you measure one observable but time dependent when you measure another observable. $\endgroup$
    – gandalf61
    Sep 25, 2019 at 11:50
  • $\begingroup$ @gandalf61 So are you implying that a quantum algorithm has to specify exact times (not just order) at which operations are performed? I.e. suppose my aparatus allows me to prepare a qubit in state $|+\rangle$ and measure in the basis $B$ consistintg of $|+\rangle$ and $|-\rangle$. Then are you saying in particular that the following is not an algorithm: 1) prepare a qubit in the state $|+\rangle$, 2) measure in the basis $B$. If this is not a quantum algorithm then I would really appreciate some reference about what constitutes a quantum algorithm. $\endgroup$ Sep 25, 2019 at 11:56

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